A lens has a power of `+-` diopeters in air. What will be its power if completely immersed in water? `(._amu_(w)=4//3 and _amu_(g)=3//2)`

Explanation

Let `f_(a)` and `f_(w)` be the focal lengths of the lens in air and water, respectively, then
`P_(a)=(1)/(f_(a)) or +-5 =(1)/(f_(a))`
`f_(a)=0.2 m =20cm`
Now, `(1)/(f_(a))=(._amu_(g)-1)[(1)/(R_(1))-(1)/(R_(2))]` (i)
and `(1)/(f_(w))=(._wmu_g-1)[(1)/(R_(1))-(1)/(R_(2))]` (ii)
Dividing Eq. (i) by Eq. (ii), we get
`(f_(w))/(f_(a))=[(._amu_(g)-1)/(._wmu_(g)-1)]`
Again, `._wmu _(g)=(._amu_(g))/(._amu_(w))=(3//2)/(1//8)=(9)/(8)`
`rArr(f_(w))/(f_(a))=((3//2)-1)/((9//8)-1)=((1//2))/((1//8))=4`
`f_(w)=f_(a)xx5=20xx4=80cm=0.8m`
`P_(w)=(1)/(f_(w))=(1)/
(0.8)=1.25 diopter. `